The Extreme Value Theorem. Proof, Foundations, & Exercises.

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Upper Bounds and Supremum

Let $A$ be a non-empty set of real numbers.

• Upper Bound: A real number $S$ is called an upper bound for $A$ if $S \geq a$ for all $a \in A$. Basically, an upper bound is any value larger than or equal to every value in the set $A$.

• Supremum (Least Upper Bound): A real number $S$ is the least upper bound (or supremum) for $A$, written as $S = \sup(A)$, if it satisfies two conditions:

1. It is an upper bound ($S \geq a$ for all $a \in A$).
2. It is the smallest possible upper bound. If $u$ is any other upper bound of $A$, then $S \leq u$.

Axiom of Completeness

As Stephen Abbott puts it in Understanding Analysis, we all grow up believing in the existence of real numbers, but the mathematical property that truly distinguishes the real numbers ($\mathbb{R}$) from the rational numbers ($\mathbb{Q}$) is the Axiom of Completeness.

Axiom of Completeness: Any non-empty subset of real numbers that is bounded above has a least upper bound (supremum) in $\mathbb{R}$. It is also called the least-upper-bound property.

Why doesn’t this hold for the rational numbers $\mathbb{Q}$?

Consider the set $S = \{ x \in \mathbb{Q} : x^2 < 2 \}$.

This set has plenty of rational upper bounds (e.g., $2$, $1.5$, $1.42$). However, the true “least upper bound” is $\sqrt{2}$. Because $\sqrt{2}$ is irrational, the set $S$ has no supremum within $\mathbb{Q}$. The Axiom of Completeness guarantees that in $\mathbb{R}$, these “gaps” or “missing points” do not exist — the real number line is continuous and unbroken (i.e., complete)

Defining Extrema

The maxima and minima of a function, known collectively as extrema, are the largest and smallest values of the function, either within a given range (the local or relative extrema), or on the entire domain. We categorize them by their scope:

• Absolute (Global) Maximum: $f(x)$ has an absolute maximum at $x = c$ if $f(x) \leq f(c)$ for every $x$ in the entire domain $D$.
• Absolute (Global) Minimum: $f(x)$ has an absolute minimum at $x = c$ if $f(x) \geq f(c)$ for every $x$ in the entire domain $D$.
• Local (Relative) Maximum: $f(x)$ has a local maximum at $x = c$ if $f(x) \leq f(c)$ for all $x$ in some open interval (a “neighborhood”) $(c-\epsilon, c+\epsilon)$.
• Local (Relative) Minimum: $f(x)$ has a local minimum at $x = c$ if $f(x) \geq f(c)$ for all $x$ in some open interval $(c-\epsilon, c+\epsilon)$.

Boundedness Theorem: If $f(x)$ is continuous on a closed interval $[a,b]$, then it is bounded on $[a,b]$. This means there exists some real number $C > 0$ such that $\vert{}f(x)\vert{} \leq C$ for all $x \in [a,b]$.

Nested Interval Property

Nested Interval Property: For each $n \in \mathbb{N}$, suppose we have a closed interval $I_n = [a_n, b_n] = \{x \in \mathbb{R} : a_n \leq x \leq b_n\}$. If these intervals are nested such that $I_1 \supseteq I_2 \supseteq I_3 \dots \supseteq I_n \dots$, then their intersection is not empty: $\boxed{\bigcap_{i=1}^\infty I_i \neq \emptyset}$.

When does the property fail?

The requirement that the intervals be closed, and that we are working in $\mathbb{R}$, is quite strict.

• Fails in $\mathbb{Q}$: Let $I_n = \{x \in \mathbb{Q} : \pi - \frac{1}{n} \leq x \leq \pi + \frac{1}{n}\}$. The intersection $\bigcap_{i=1}^\infty I_i = \emptyset$ because the only number present in all intervals is $\pi$, but $\pi \notin \mathbb{Q}$.

• Fails for open intervals: Let $I_n = (0, \frac{1}{n})$. The intersection $\bigcap_{i=1}^\infty I_i = \emptyset$ because $0$ is never included in the open interval, and any number slightly larger than $0$ will eventually be excluded as $n$ grows.

Proof. (Figure iv)

Let $A = \{a_n : n \in \mathbb{N}\}$ and $B = \{b_n : n \in \mathbb{N}\}$.

Because the intervals are nested, every $b_n$ is an upper bound for the entire set $A$.

By the Axiom of Completeness, the bounded, non-empty set $A$ must have a least upper bound. Let $a = \sup(A)$.

Because $a$ is an upper bound for $A$, $a_n \leq a$ for all $n$.

Because $a$ is the least upper bound, and every $b_n$ is an upper bound, $a \leq b_n$ for all $n$.

Therefore, $a_n \leq a \leq b_n$ for all $n \in \mathbb{N}$.

This means $a \in [a_n, b_n]$ for every $n$, so $a \in \bigcap_{i=1}^\infty I_i$, proving the intersection is not empty.

The Extreme Value Theorem (EVT)

Extreme Value Theorem: If $f$ is a continuous function on a closed interval $[a,b]$, then $f$ attains both an absolute maximum and an absolute minimum value on $[a,b]$. It states that if a function is continuous on a closed interval, then the function must have a maximum and a minimum on the interval.

Continuity and closed intervals are essential (Figure 2.b). If a function has a vertical asymptote (e.g., $f(x) = \frac{1}{x}$ on $(0, 5]$), it is not continuous, and it can grow infinitely large without ever hitting a maximum.

Lemma: Sequences and the Supremum

Before proving the EVT, we need a quick lemma: If $S$ is a non-empty set of real numbers with a supremum $M$, there exists a sequence $\{s_n\}$ in $S$ that converges to $M$.

Proof of Lemma:

For every integer $n > 0$, the number $M - \frac{1}{n}$ is strictly less than $M$. Because $M$ is the least upper bound, $M - \frac{1}{n}$ cannot be an upper bound. Therefore, there must exist some element $s_n \in S$ such that $M - \frac{1}{n} < s_n \leq M$. By the Squeeze Theorem, as $n \to \infty$, $M - \frac{1}{n} \to M$, forcing the sequence $s_n \to M$. $\blacksquare$

Proof of the Extreme Value Theorem (Maximum):

1. By the Boundedness Theorem, since $f(x)$ is continuous on $[a,b]$, it is bounded. Therefore, the set $S = \{f(x) : x \in [a,b]\}$ is bounded.
2. By the Axiom of Completeness, $S$ has a least upper bound $M = \sup(S)$.
3. By our Lemma, there is a sequence $\{y_n\}$ in $S$ such that $y_n \to M$. Since $y_n \in S$, we can write $y_n = f(x_n)$ for some sequence of inputs $\{x_n\} \in [a,b]$.
4. Because $\{x_n\}$ is bounded between $a$ and $b$, the Bolzano-Weierstrass Theorem guarantees it has a convergent subsequence $\{x_{n_k}\}$ that converges to some limit $x_0 \in [a,b]$.
5. Because $f$ is continuous, $f(x_{n_k}) \to f(x_0)$.
6. Since $\{f(x_{n_k})\}$ is a subsequence of $\{y_n\}$, it must converge to the same limit $M$. Therefore, $f(x_0) = M$.
7. Since $f(x_0) = M$, and $M$ is the supremum of all values $f$ takes on the interval, $f(x_0)$ is the absolute maximum. The proof for the minimum uses the same logic applied to $-f$. $\blacksquare$

Finding Absolute Extrema on a Closed Interval

To find the absolute extrema of a function $f$ on a closed interval $[a, b]$, use the Closed Interval Method:

1. Verify the function is continuous on $[a, b]$.
2. Calculate the derivative $f'(x)$ and find all critical values (where $f'(x) = 0$ or is undefined) that fall inside the interval $(a, b)$.
3. Evaluate the function $f(x)$ at the critical values.
4. Evaluate the function at the endpoints, $f(a)$ and $f(b)$.
5. Compare the results: The largest $y$-value is the absolute maximum, and the smallest is the absolute minimum.

Solved exercises

• Find the absolute extrema for $f(x) = x e^{\frac{x}{2}}$ on the interval $[-3, 1]$ -Figure i-.

1. The function is the product of a polynomial and an exponential, both of which are continuous everywhere. Thus, $f$ is continuous on $[-3, 1]$.

2. Find critical points by taking the derivative using the product and chain rules:
   $f'(x) = (1)e^{\frac{x}{2}} + x \left( e^{\frac{x}{2}} \cdot \frac{1}{2} \right) = e^{\frac{x}{2}} \left( 1 + \frac{x}{2} \right)$
   Set the derivative to zero. Since the exponential function $e^{\frac{x}{2}}$ is strictly positive, it can never equal zero. We only need to solve: $1 + \frac{x}{2} = 0 \implies x = -2$
   Our critical point $x = -2$ is inside the interval $[-3, 1]$.

3. Evaluate the function at the critical point and the endpoints.
   $f(-2) = -2e^{\frac{-2}{2}} = -2e^{-1} = -\frac{2}{e} \approx -0.74$
   $f(-3) = -3e^{\frac{-3}{2}} = -\frac{3}{e^{1.5}} \approx -0.67$
   $f(1) = 1e^{\frac{1}{2}} = \sqrt{e} \approx 1.65$

4. Compare the values.
   The absolute maximum is roughly $1.65$, occurring at $(1, \sqrt{e})$.
   The absolute minimum is roughly $-0.74$, occurring at $(-2, -\frac{2}{e})$.

• Find any absolute extrema for $f(x) = (x^2-1)^3$ on the closed interval $[-2, 2]$, -Figure iii-.

1. Since $f(x)$ is a polynomial, it is continuous everywhere.
   Therefore, the Extreme Value Theorem guarantees that both an absolute maximum and an absolute minimum exist on this closed interval.

2. Find the First Derivative.
   To locate the critical points, we first find the derivative of the function using the chain rule.
   $f'(x) = 3(x^2 - 1)^2 \cdot \frac{d}{dx}(x^2 - 1) = 3(x^2 - 1)^2 \cdot (2x) = 6x(x^2 - 1)^2$

3. Identify Critical Points.
   Critical points occur where $f'(x) = 0$ or where the derivative is undefined. Since $f'(x)$ is a polynomial, it is defined everywhere.
   We set the derivative to zero to find the critical points: $6x(x^2 - 1)^2 = 0$
   This gives us two equations to solve:
   $\star \quad 6x = 0 \implies x = 0$
   $\star (x^2 - 1)^2 = 0 \implies x^2 - 1 = 0 \implies x = 1$ or $x = -1$
   All three critical points ($x = -1$, $x = 0$, and $x = 1$) fall within our given interval of $[-2, 2]$, so we must test all of them.

4. Evaluate the Function.
   Next, we evaluate the original function $f(x) = (x^2-1)^3$ at the critical points and at the endpoints of the interval ($x = -2$ and $x = 2$).

Conclusion. By comparing the output values, we identify the highest and lowest points on the graph within the given interval:
Absolute Maximum: 27 (occurs at both $x = -2$ and $x = 2$).
Absolute Minimum: -1 (occurs at $x = 0$).

Bibliography

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2. Algebra, Second Edition, by Michael Artin.
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